The maximum value of sin x . cos x is

The correct option is 1/2.

To find the maximum value of the expression $f\left(x\right)=sinx·cosx$, we can simplify it using trigonometric identities.

Recall the double-angle identity for sine:
$sin\left(2x\right)=2sinxcosx$

By dividing both sides of this identity by 2, we can rewrite the original expression as:
$sinx·cosx=\frac{1}{2}sin\left(2x\right)$

Now, we analyze the range of the sine function. For any real angle $\theta$, the value of $sin\theta$ always lies between -1 and 1, inclusive:
$-1\le sin\left(2x\right)\le 1$

To find the range of $\frac{1}{2}sin\left(2x\right)$, we multiply the entire inequality by $\frac{1}{2}$:
$-\frac{1}{2}\le \frac{1}{2}sin\left(2x\right)\le \frac{1}{2}$

Thus, the maximum value that the expression $sinx·cosx$ can achieve is $\frac{1}{2}$, which occurs when $sin\left(2x\right)=1$ (for example, when $x=\frac{\pi }{4}$).